Ruby on rails 找到最小值'；配对'；排列

我有一个“数组对”数组：

[[time, distance], [time, distance], [time, distance], [time, distance], ...]

我想找到具有最小时间的“对”的索引，如果有多个具有相同最小时间的“对”，则取具有最小距离的对

e、 g.如果我有：

[[5, 5], [1,7], [2,6], [1,6]]

我想返回3。（即第四个元素的索引）

我该怎么做呢？

Array#sort

正是您想要的

myarray = [[5, 5], [1,7], [2,6], [1,6]]
myarray.sort
#=> [[1, 6], [1, 7], [2, 6], [5, 5]]

因此，我们只需要找到原始数组中与排序数组中的第一个元素匹配的元素

ndx = myarray.index myarray.sort[0]
#=> 3

---

编辑：我最初使用spaceship操作符，但意识到它不是必需的。

我不熟悉RubyonRails，但我编写了一些PHP代码来实现您的目标。我希望你能从中提炼出这个想法

<?php

$timesAndDistances = [['t' => t1, 'd' => d1], ['t' => t2, 'd' => d2], ..., ['t' => tn, 'd' => dn]];

$lowestTime_withDistance = [['t' => null, 'd' => null, 'i' => null]];

foreach($timesAndDistances as $index => $timeAndDistance)
{
    $time = $timeAndDistance['t'];
    $distance = $timeAndDistance['d'];

    if($lowestTime_withDistance[0]['t'] == null)
    {
        $lowestTime_withDistance[0] = ['t' => $time, 'd' => $distance, 'i' => $index];
    }
    else 
    {
        if($lowestTime_withDistance[0]['t'] > $time){
            $lowestTime_withDistance[0] = ['t' => $time, 'd' => $distance, 'i' => $index];
        }
        elseif ($lowestTime_withDistance[0]['t'] == $time)
        {
            $lowestTime_withDistance[] = ['t' => $time, 'd' => $distance, 'i' => $index];
        }
    }
}

if(count($lowestTime_withDistance) == 1)
{
    echo 'Index of element with minimum time is ' . $lowestTime_withDistance[0]['i'];
}
elseif(count($lowestTime_withDistance) > 1){
    $lowestDistance = [['d' => null, 'i' =>null]];

    foreach($lowestTime_withDistance as $timeWithDistance)
    {
        if($lowestDistance[0]['d'] == null)
        {
            $lowestDistance[0] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']];
        }
        else
        {
            if($lowestDistance[0]['d'] > $timeWithDistance['d'])
            {
                $lowestDistance[0] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']];
            }
            elseif ($lowestDistance[0]['d'] == $timeWithDistance['d'])
            {
                $lowestDistance[] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']];
            }
        }
    }

    if(count($lowestDistance) == 1)
    {
        echo 'Index of element with minimum time and distance is ' . $lowestDistance[0]['i'];
    }
    elseif (count($lowestDistance) > 1)
    {
        foreach($lowestDistance as $aLowestDistance)
        {
            echo 'Index of an element with minimum time and distance is ' . $aLowestDistance['i'];
        }
    }
}

---

编写快速代码以实现此目的：

假设数组a有以下元素

[5,5]，[1,7]，[2,6]，[1,6]]


在b中创建阵列的副本：

b=a

b=[[5,5]，[1,7]，[2,6]，[1,6]]


b类

b=b.排序

b=[[1,6]，[1,7]，[2,6]，[5,5]]


迭代嵌套循环

最小值=b[0]

索引=0

a、 除非是零，否则每一个都是零

如果iter_orig.eql最小

中断

结束

索引++

结束

结束

放入“最小时间/距离索引-”+索引